What is the solution for y^{''}-y^'-42y=0 ?

The solution for y^{''}-y^'-42y=0 is y=c_{1}e^{7t}+c_{2}e^{-6t}

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y^{''}-y^'-42y=0

y^{^{''}} - y^{^{'}} - 42 y = 0

y = c_{1} e^{7 t} + c_{2} e^{- 6 t}

Solution steps

y^{''} (t) - y^{'} (t) - 42 y = 0

Solve linear ODE:

y = c_{1} e^{7 t} + c_{2} e^{- 6 t}

y = c_{1} e^{7 t} + c_{2} e^{- 6 t}

d er i v a t i v e \frac{x ^{2}}{2 - x ^{2}}

\int x sin (- 1 x) d x

\int e^{- y} d y

\frac{d}{d x} (ln (3 x + 1))

x \to 0 lim (\frac{1 - 2 cos ( x ) + cos ( 2 x )}{x ^{2}})

What is the solution for y^{''}-y^'-42y=0 ?
The solution for y^{''}-y^'-42y=0 is y=c_{1}e^{7t}+c_{2}e^{-6t}